# The Plucker relations are sufficient

Consider the Grassmannian of codimension-$d$ subspaces of a given vector space $E$ (over an arbitrary field), which I will define as $$\operatorname{Gr}^d(E) = \{\text{linear surjections } \sigma: E \to F \mid \text{F is any d-dimensional space} \}/\sim$$ where I identify two surjections $(\sigma_1: E \to F_1) \sim (\sigma_2: E \to F_2)$ if there is an isomorphism $m: F_1 \to F_2$ such that $m \sigma_1 = \sigma_2$. Then an isomorphism class of $\sigma$ is determined by the $\ker \sigma$. We also have projective space $\mathbb{P}(E) = \operatorname{Gr}^1(E)$, so that $\operatorname{Sym}^\bullet(E)$ gives homogeneous coordinates on $\mathbb{P}(E)$.

The Plucker embedding is the map $\operatorname{Gr}^d(E) \to \mathbb{P}(\bigwedge^d E)$, taking a map $(\sigma: E \to F)$ to its $d$th exterior power $\bigwedge^d \sigma: \bigwedge^d E \to \bigwedge^d F$. Then $\bigwedge^d \sigma$ is a surjection from $\bigwedge^d E$ to a one-dimensional space, and so lives in $\mathbb{P}(\bigwedge^d E)$. So I am thinking of $\bigwedge^d \sigma$ as a point in the embedding of the Grassmannian, and elements of $\bigwedge^d(E)$ give me the linear coordinate functions. For example, if I write out the matrix of $\sigma: E \to F$ in some basis $(e_1, \ldots, e_m)$ for $E$ and any basis for $F$, then $(\bigwedge^d \sigma)(e_1 \wedge \cdots \wedge e_d)$ is proportional to the determinant of minor where we take the first $d$ columns of the matrix for $\sigma$.

Of course, the question arises that, given some $(s: \bigwedge^d E \to L) \in \mathbb{P}(\bigwedge^d E)$, is $s$ in the image of the Plucker embedding? (i.e., is $s$ a $d$th wedge power?). The answer to this is given by the Plucker relations, which I will state as follows. Define a linear map $\omega^d_E$ on pure tensors by \begin{aligned}\omega^d_E: \bigwedge^{d+1} E \otimes \bigwedge^{d-1} E &\to \bigwedge^d E \otimes \bigwedge^d E\\ v_1 \wedge \cdots \wedge v_{d+1} \otimes u_1 \wedge \cdots \wedge u_{d-1} &\mapsto \sum_{i = 0}^{d+1} v_1 \wedge \cdots \wedge \hat{v_i} \wedge \cdots \wedge v_{d+1} \otimes v_i \wedge u_1 \wedge \cdots \wedge u_{d-1} \end{aligned} Then the point $(s: \bigwedge^d E \to L)$ "satisfies the Plucker relations" if the linear map $(s \otimes s) \circ \omega^d_E$ is zero. From this standpoint, it is lovely to see the necessary condition: if $(\sigma: E \to F) \in \operatorname{Gr}^d(E)$, then $\wedge^d \sigma$ must satisfy the Plucker relations, since the wedge power essentially "pulls through" the $\omega$: $$(\wedge^d \sigma \otimes \wedge^d \sigma) \circ \omega^d_E = \omega_F^d \circ (\wedge^{d+1} \sigma \otimes \wedge^{d-1} \sigma)$$ and of course $\wedge^{d+1} \sigma = 0$ because $F$ is $d$-dimensional. However, I am having trouble with the opposite direction.

How can I see that if $(s: \wedge^d E \to L) \in \mathbb{P}(\bigwedge^d E)$ satisfies the Plucker relations, then it is (proportional to) the $d$th power of some map $\sigma: E \to F$? I would really like a constructive proof: produce some $\sigma: E \to F$ from $s$, and show that as long as $s$ satisfies the Plucker relations, that $\wedge^d \sigma \sim s$.

I can prove similar results for decomposable tensors in $\wedge^d E$, but for some reason with how things are set up, "dualising" that proof is eluding me. There is a proof similar to the style I am thinking of on page 175 of Martin Brandenburg's Thesis, unfortunately I cannot follow the proof starting from the "two short exact sequences". Which leads me to another question:

Are there any good references for the Plucker embedding or Plucker relations thought of in this way?

• What exactly don't you understand in Brandenburg's proof? In the two exact sequences, the left-most map of the first row takes $a\otimes b$ to $a\otimes s(b)-b\otimes s(a)$ and the left-most map of the bottom is just the natural map $\ker t\otimes \bigwedge^{d-1}\to \bigwedge E$, twisted by $L$. (Note that $\ker t = \bigwedge^{d+1}E\otimes L^\vee$ by construction.) That the two diagonal compositions vanish imply that the right-most square can be completed by an isomorphism. Twisting by $L^\vee$ again gives exactly what you want. – Ben May 4 '18 at 17:03
• @Ben Literally everything you just said was not clear to me. How do you know that the upper left map takes $a \otimes b$ to $a \otimes s(b) - b \otimes s(a)$ rather than $a \otimes s(b)$ for example? Also where can I find this result about diagonals vanishing in two short exact sequences? I guess the other thing is that this proof seems entirely pulled-out-of-a-hat, and I'm trying to get some intuition for exactly what is going on. I understand why defining $t$ as the cokernel of a certain map should be right, but I cannot make any sense of the proof from there. – Joppy May 5 '18 at 0:46
• I see. As for intuition, I don’t know, but I will try to explain the details there a bit further later. – Ben May 5 '18 at 6:37
• @Ben Thanks, that would be much appreciated. – Joppy May 5 '18 at 6:41

As promised in the comment, here are some more details to Martin Brandenburg's proof. To get an understanding of what is going on, we first put ourselves in the situation that $s = \wedge^d\sigma$ for some surjective $\sigma\colon E\to F$. Why would we want to consider the map $t\colon \bigwedge\nolimits^{d+1}E\to E\otimes\bigwedge\nolimits^{d}F$ defined as \begin{align*}t(v_0\wedge\dots\wedge v_d) &= \sum_{k=0}^d(-1)^kv_k\otimes s(v_0\wedge\dots\wedge\widehat{v_k}\wedge\dots\wedge v_d) \\&=\sum_{k=0}^d(-1)^kv_k\otimes \sigma(v_0)\wedge\dots\wedge\widehat{\sigma(v_k)}\wedge\dots\wedge \sigma(v_d)\;\;? \end{align*} Well, I claim that its image is the kernel of the map $E\otimes\bigwedge^dF\xrightarrow{\sigma\otimes \mathrm{id}_{\wedge^d F}} F\otimes\bigwedge^dF$; since we can reconstruct $\sigma\colon E\to F$ up to isomorphism from its kernel, this map is very much relevant for what we are trying to do.

Let's prove the claim: It is easy to verify that $(\sigma\otimes \mathrm{id}_{\wedge^dF})\circ t$ factorises as $$\bigwedge\nolimits^{d+1}E\xrightarrow{\wedge^{d+1}\sigma}\bigwedge\nolimits^{d+1}F\to F\otimes\bigwedge\nolimits^{d}F,$$ where the latter map sends $v_0\wedge\dots\wedge v_d$ to $\sum_{k=0}^d(-1)^kv_k\otimes v_0\wedge\dots\wedge\widehat{v_k}\wedge\dots\wedge v_d$. But $\bigwedge\nolimits^{d+1}F = 0$; thus, $(\sigma\otimes\mathrm{id}_{\wedge^dF})\circ t = 0$, so that $\mathrm{im}(t)\subset \ker(\sigma\otimes \mathrm{id}_{\wedge^dF})$. Conversely, if $\sigma(w) = 0$, then $t(w\wedge v_1\wedge\dots\wedge v_d)=w\otimes s(v_1\wedge\dots\wedge v_d)$ and so $t$ maps surjectively onto $\ker(\sigma)\otimes\bigwedge^dF = \ker(\sigma\otimes \mathrm{id}_{\wedge^dF})$, as claimed.

What this means is that we have found a reasonable candidate for an inverse of the map $$\left\{E\xrightarrow{\sigma} F\to 0\right\}\to\left\{\bigwedge\nolimits^{d}E\xrightarrow{s}L\to 0\,\middle|\,\text{sat. Plücker}\right\},\sigma\mapsto \wedge^d\sigma,$$ by mapping $s$ to the cokernel of $T_s\colon \bigwedge\nolimits^{d+1}E\otimes L^\vee\xrightarrow{t\otimes \mathrm{id}_{L}}E\otimes L\otimes L^\vee\to E$, where the last map is just the natural isomorphism. The above shows that if we start with a $\sigma$, pass to $\wedge^d\sigma$, and then take the cokernel of $T_{\wedge^d\sigma}$, we get back $\sigma$ up to isomorphism. It remains to show that starting with some $s$ satisfying the Plücker relations, the candidate-inverse is well-defined (i.e., that the cokernel has rank $d$,) and that if we pass to the cokernel $\sigma\colon E\to F:=\mathrm{coker}(T_s)$ and then apply $\wedge^d$, we get back $s$ up to isomorphism. The latter is what those exact sequences are for, but we can phrase it without them:

We have two quotients of $\bigwedge^dE\otimes L$, namely, $\wedge^d\sigma\otimes \mathrm{id}_{L}\colon \bigwedge^dE\otimes L\to \bigwedge^dF\otimes L$ and $s\otimes \mathrm{id}_{L}\colon\bigwedge^dE\otimes L\to L\otimes L$ and we aim to show that they are isomorphic as quotients, i.e., that $\ker(\wedge^d\sigma\otimes\mathrm{id}_{L}) = \ker(s\otimes \mathrm{id}_{L})$. For this, we give nice presentations of those kernels.

For one, since $\ker(\sigma\otimes\mathrm{id}_{L})$ is the image of $t$, the kernel of $\wedge^d\sigma\otimes\mathrm{id}_{L}$ is the image of the map $\alpha\colon \bigwedge^{d-1}E\otimes\bigwedge^{d+1}E\to \bigwedge^{d}E\otimes L$, mapping $v\otimes w$ to $v\wedge t(w)$.

For the other map, note that $\ker(s\otimes \mathrm{id}_{L})=\ker(s)\otimes L$ is generated by elements of the form $v\otimes s(w)-w\otimes s(v)$, since, for $s(v)=0$ and $f = s(w)\in L$ arbitrary, $v\otimes s(w) - w\otimes s(v) = v\otimes f$. In particular, with $\beta\colon \bigwedge^dE\otimes\bigwedge^dE\to \bigwedge^dE\otimes L$ mapping $v\otimes w$ to $v\otimes s(w)- w\otimes s(v)$, we get $\ker(s\otimes \mathrm{id}_{L}) = \mathrm{im}{(\beta)}$.

Thus, if we manage to show that $(s\otimes \mathrm{id}_{L})\circ\alpha = 0$ and $(\wedge^d\sigma\otimes\mathrm{id}_{L})\circ\beta = 0$, then we conclude $$\ker(\wedge^d\sigma\otimes\mathrm{id}_{L})=\mathrm{im}{(\alpha)}\subset\ker(s\otimes \mathrm{id}_{L}) = \mathrm{im}{(\beta)}\subset \ker(\wedge^d\sigma\otimes\mathrm{id}_{L}),$$ which implies equality everywhere. In particular, $L\cong\bigwedge^dF$ as quotients of $\bigwedge^d E$ and so $\bigwedge^dF$ is invertible, hence $F$ has rank $d$; this is all we wanted to show.

Finally, we show the two identities $(s\otimes \mathrm{id}_{L})\circ\alpha = 0$ and $(\wedge^d\sigma\otimes\mathrm{id}_{L})\circ\beta = 0$. Tracing through the definitions shows that the former is the Plücker relation, and that the second is equivalent to $\wedge^d\sigma v\otimes s(w) = \wedge^d\sigma w\otimes s(v)$ for all $v,w\in\bigwedge^dE$. That is, we want $\wedge^d\sigma\otimes s$ to be symmetric. By construction of $\sigma$, we always have $$0 = \sum_{k=0}^d(-1)^{k}\sigma w_k\otimes s(w_0\wedge\dots\wedge \widehat{w_{k}}\wedge \dots\wedge w_d),$$ and so the symmetry of $\wedge^d\sigma\otimes s$ follows from what M. Brandenburg calls the Symmetry Lemma (4.4.15); I have nothing to add to his proof of this lemma.

• Thanks for taking the time to explain! I still don't understand how you go from the fact that the kernel of $\sigma \otimes \mathrm{id}_L$ is the image of $t$, and arrive at the conclusion that the kernel of $\wedge^d \sigma \otimes \mathrm{id}_L$ is precisely the image of $\alpha$. – Joppy May 8 '18 at 1:27
• This holds quite generally: if $\varphi\colon V\to W$ is surjective, then so is $\wedge^d \varphi$ and the kernel is generated by the elements of the form $v_1\wedge\dots\wedge v_d$ for all $v_1\in\ker\varphi$ and $v_2,\dots,v_d\in V$. I will try to find a reference or a quick argument later in the day. – Ben May 8 '18 at 5:31
• Ah, I see - I definitely believe that. I think I'm beginning to see how to connect this to the usual decomposability of vectors in $\wedge^d E$ results. – Joppy May 8 '18 at 6:29
• Here is a simple argument: If $\wedge^d\varphi(v_1,\dots,v_d) = 0$, then the $\varphi(v_i)$ are linearly dependent. Thus, there exist scalars $\lambda_i$ such that $\varphi(\sum_i\lambda_i v_i) = \sum_i\lambda_i\varphi(v_i) = 0$, hence, $v_0 := \sum_i\lambda_i v_i\in\ker(\varphi)$. Say $\lambda_1 = 1$, possibly after renumbering and rescaling. Then $v_1\wedge\dots\wedge v_d = v_0\wedge v_2\wedge\dots\wedge v_d - (v_0-v_1)\wedge v_2\wedge\dots\wedge v_d$ where the first term has a factor in $\ker(\varphi)$ and the second vanishes. It remains to show that it suffices to consider those elements.. – Ben May 8 '18 at 7:05

I cannot put this as a comment and I'm sorry that it is not a complete answer, but a good introduction to Grassmanninans is written by Gathmann: http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2014/alggeom-2014-c8.pdf

In particular the answer to your second question could be Corollary 8.13 (look at the proof).

• Thanks for the reference, but this isn't what I'm looking for. In those notes, the relations given are the vanishing of some $(n-1+k) \times (n-1+k)$: this is a relation of degree $n-1+k$. The Plucker relations I gave above are different, and in particular are always degree 2. – Joppy Apr 29 '18 at 1:36